package editor.cn;//给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
//
// 给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。
//
//
//
// 示例:
//
// 输入："23"
//输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
//
//
// 说明:
//尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。
// Related Topics 字符串 回溯算法
// 👍 882 👎 0

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class LetterCombinationsOfAPhoneNumber {
    public static void main(String[] args) {
        Solution solution = new LetterCombinationsOfAPhoneNumber().new Solution();
        solution.letterCombinations("23");

    }

    class Solution {
        public List<String> letterCombinations(String digits) {
            List<String> combinations = new ArrayList<String>();
            if (digits.length() == 0) {
                return combinations;
            }
            Map<Character, String> phoneMap = new HashMap<Character, String>() {{
                put('2', "abc");
                put('3', "def");
                put('4', "ghi");
                put('5', "jkl");
                put('6', "mno");
                put('7', "pqrs");
                put('8', "tuv");
                put('9', "wxyz");
            }};
            backtrack(combinations, phoneMap, digits, 0, new StringBuffer());
            System.out.println(combinations);
            return combinations;
        }

        public void backtrack(List<String> combinations, Map<Character, String> phoneMap, String digits, int index, StringBuffer combination) {
            //1 @index = 0 digits=23
            //2 @index = 1 digits=23
            //3 @index = 2 digits=23
            if (index == digits.length()) {
                combinations.add(combination.toString());
            } else {
                //1 @digits =23 index=0   #digits.charAt(0)==2     charAt() 方法用于返回指定索引处的字符。索引范围为从 0 到 length() - 1。
                //2 @digits =23   #digits.charAt(1)==3
                char digit = digits.charAt(index);
                //1  phoneMap.get(2)==abc
                //2  phoneMap.get(3)==def
                String letters = phoneMap.get(digit);
                //1  @letters=abc  #lettersCount==3
                //2  @letters=def  #lettersCount==3
                int lettersCount = letters.length();

                for (int i = 0; i < lettersCount; i++) {
                    //1 @letters.charAt(i) i=0  # combination==a
                    combination.append(letters.charAt(i));
                    //1      0 8 23 1 a
                    backtrack(combinations, phoneMap, digits, index + 1, combination);
                    combination.deleteCharAt(index);
                }
            }
        }
    }

}